I cant follow the logic of this proof

[transgalactic]

W_1 and W_ 2 are subspaces of V of inner product V.
prove that if
$$W_1\subseteq W_2$$
then
$$W_1^\perp \supseteq W_2^\perp$$

the proof is:
we take $$v\epsilon W_1$$
so <v,w>=0 for every $$w\epsilon W_1^\perp$$

and because
$$W_2^\perp \subseteq W_1^\perp$$
(i can't see why the "viven expression is "true" why "because")
we get that
<v,w>=0
for every $$w\epsilon W_2^\perp$$

so $$v\epsilon W_2$$
so
$$W_1^\perp \supseteq W_2^\perp$$

??

 

[Office_Shredder]

Your latex is messed up a bit, but it looks like what you typed after "because" is intended to be the final conclusion, which doesn't make a whole lot of sense. What you want is

<v,w>=0 for all w in W₂ then <v,w> = 0 for all w in W₁ (because W₁ is a subset) Hence v in W₂^perp implies v is in W₁^perp by the definition of the perpendicular subspace

 

[transgalactic]

sorry i ment
$$<br /> W_2^\perp \subseteq W_1^\perp <br />$$i can't see how they came to this conclusion after they defined v
??

 

[transgalactic]

we got two subspaces W1 and W2
W2 includes W1
they take "v" which is orthogonal to every vector which is orthogonal to W1

so v is parallel to W1 ,v is a part of W1 .

you say
"<v,w>=0 for all w in W2"
so all the vectors in w (W2) are perpendicular to v (which is W1)

no one told us that W2 is perpendicular to W1 ??

further more i was told that W2 include W1

W2 W1 cannot be orthogonal
??

 

[HallsofIvy]

The way to prove A is a subset of B is to start "if x is a member of A" and conclude "x is a member of B".
If x is a member of $$W_2^\perp$$ then <x, v>= 0 for every member, v, of $$W_2$$. Since $$W_1$$ is a subset of $$W_2$$, every v in $$W_1$$ is a member of $$W_2$$ so <x, v>= 0 for every member of $$W_1$$.

 

[transgalactic]

ok i agree <v,x>=0 because W1 is a part of W2

what is the next logical step
its still doesn't give me the step i didnt understang

$$W_2^\perp \subseteq W_1^\perp$$
how did they get it?

 

[transgalactic]

v is perpendicular to $$W_1^\perp$$
v is perpendicular to $$W_2^\perp$$ because W1 is a part of W2
so
one perpendicular member is a part of the other perpendicular member
but how to decide W2 perpendicular is a part of W1 perpendicular

$$W_2^\perp \subseteq W_1^\perp$$
why its not the other way around
??

 

[transgalactic]

how to deside?

 

[Mark44]

Maybe a diagram would help. Here's a representation of the vector space V and the two subspaces, $W_1 and W_2$, with $W_1 \subseteq W_2$.



$W_2^\perp$ is everything outside of $W_2$, but including the 0 vector.
$W_1^\perp$ is everything outside of $W_1$, which means it includes everything in $W_2$ that's not also in $W_1$, plus everything that's outside of $W_2$, plus the 0 vector.

Now, any vector $v \in W_1^\perp$ must be in
$(W_2 - W_1) \cup \{0\} \cup W_2^\perp$

$\forall v \in W_2^\perp$,
$v \in W_2^\perp \cup (W_2 - W_1) \cup \{0\} = W_1^\perp$.
Therefore, $W_2^\perp \subseteq W_1^\perp$

 

[transgalactic]

the diagram is pending approval can you post a link
from
image shack

http://imageshack.us/
??

 

[arildno]

We know that every perpendicular vector of the larger subspace W_2 must be perpendicular to all vectors of any subspace within W_2; including the vectors making up W_1.

Thus, ALL vectors in the perpendicularity space of W_2, are elements in the perpendicularity space of W_1.
This means that the perp space of W_2 is contained in that of W_1

But, does it thereby follow that the perpendicularity space of W_1 is exhausted?
NO!

It might well include vectors of W_2 that happens to be perpendicular to every vector in W_1 as well!

Such vectors would NOT be part of the perpindicularity space of W_2, and therefore, one cannot say that the perp space of W_1 is contained within that of W_2

 

[transgalactic]

Maybe a diagram would help. Here's a representation of the vector space V and the two subspaces, $W_1 and W_2$, with $W_1 \subseteq W_2$.



$W_2^\perp$ is everything outside of $W_2$, but including the 0 vector.
$W_1^\perp$ is everything outside of $W_1$, which means it includes everything in $W_2$ that's not also in $W_1$, plus everything that's outside of $W_2$, plus the 0 vector.

Now, any vector $v \in W_1^\perp$ must be in
$(W_2 - W_1) \cup \{0\} \cup W_2^\perp$

$\forall v \in W_2^\perp$,
$v \in W_2^\perp \cup (W_2 - W_1) \cup \{0\} = W_1^\perp$.
Therefore, $W_2^\perp \subseteq W_1^\perp$

i can't see the attached diagram

 

[transgalactic]

"Thus, ALL vectors in the perpendicularity space of W_2, are elements in the perpendicularity space of W_1."
why its not the other way around

 

[arildno]

"Thus, ALL vectors in the perpendicularity space of W_2, are elements in the perpendicularity space of W_1."
why its not the other way around

Because a non-zero vector in W_2 that is not in W_1 might be perpendicular to all vectors in W_1.

Since it by definition would be IN W_2, it cannot be perpendicular to every vector in W_2 (it is, for example, NOT perpendicular to itself!).

Thus, such a vector would be in the perpendicularity space of W_1, but not in the perpendicularity space of W_2

 

[Mark44]

i can't see the attached diagram

You should be able to see it now.